Estimate for the fundamental solution of Laplace equation from Evans After the definition of the fundamental solution of Laplace 
equation (page 22)
$\Phi(x) = \begin{cases} -\frac{1}{2\pi} \, \log(|x|), \, & n=2, \\ \frac{1}{n \, (n-2) \, \omega_n} \, \frac{1}{|x|^{n-2}}, \, & n\geq 3, \end{cases}$
Evans gives two estimates:
(i) $|D \Phi(x)| \leq \frac{C}{|x|^{n-1}}$
(ii) $|D^2 \Phi(x)|\leq \frac{C}{|x|^n}$ 
for some constant $C>0$
I have been able to prove (i) with no difficulties, but I'm really stuck on (ii), for the case n \leq 3.

I know that $D^2 \Phi(x)$ denotes the hessian matrix of $\Phi$, but I don't understand how to compute its norm. As far as I know (I looked up on the appendix), given a matrix $A$, Evans writes
\begin{align}
|A|=(\sum_{i=1}^{n} \sum_{j=1}^n a_{ij}^2)^{1/2} \quad \star
\end{align} 
So I write in the following the $(i,j)$-th element of $D\Phi(x)$
\begin{align}
\frac{\partial}{\partial x_j} [\frac{\partial \Phi(x)}{\partial x_i}]=\frac{\partial}{\partial x_j}[v'(|x|) \frac{x_i}{|x|}]=v''(|x|) \frac{x_i x_j}{|x|^2} - v'(|x|) \frac{x_i x_j}{|x|^3}
\end{align}
Now I plug in the definition of $v(|x|)=C |x|^{1-N}$, $C$ constant, hence I have that 
\begin{align}
(D \Phi(x))_{ij}=C(1-N) \frac{x_i x_j}{|x|^{2+N}}-C \frac{x_i x_j}{|x|^{2+N}} = -N \frac{x_i x_j}{|x|^{2+N}}
\end{align}
Now I should just apply $\star$, but I don't know how to move from here
EDIT
I have that $(D \Phi(x))_{ij}=-N \frac{x_i x_j}{|x|^{2+N}}\leq -N \frac{|x|^2}{|x|^{2+N}}=-N \frac{1}{|x|^N}$
and since this bound does not depend on $(i,j)$ anymore I apply straightforward $\star$ and get 
\begin{align}
\frac{N}{|x|^N} \sqrt{\sum_{i=1}^N \sum_{j=1}^{N}}1 =\frac{N^2}{|x|^N} 
\end{align}
and hence I have the desired bound
EDIT$^2$ [17/5/2019]
After the comments with @SeverinSchraven, differentiating I end up with
\begin{align} \frac{\partial}{\partial x_i}[v'(|x|) \frac{x_i}{|x|}]= \ldots=\frac{C}{|x|^N}-N \frac{x_i^2}{|x|^{N+2}} \end{align}
and so I have, in the Hessian matrix:
\begin{equation} a_{ij}=
\begin{cases}
\frac{N}{|x|^N} \quad i \ne j \\
\frac{(C-N)}{|x|^N} \quad i =j 
\end{cases}
\end{equation}
So, using $\star$ I have:
$\sqrt{\sum_{i=1}^{N} \sum_{j=1}^{N} a_{ij}^2}= \sqrt{\sum_{i} \sum_{i} a_{ii}^2 + \frac{N^2}{|x|^N}}=\sqrt{\frac{1}{|x|^{2N}}(N^2+(N^2(C-N)^2)}=\frac{N}{|x|^N} \cdot \tilde{C}$, where $\tilde{C}=\sqrt{1+(C-N)^2}$
 A: After the comments, I'll write an answer.
Evans defines $|A|$, where $A$ is a matrix, as 
\begin{align}
|A|=(\sum_{i=1}^{n} \sum_{j=1}^n a_{ij}^2)^{1/2} \quad \star
\end{align}
so, in order to compute the norm of $D^2 \Phi$ (Hessian matrix), I need to compute each component $a_{ij}=\frac{\partial}{\partial x_i}(\frac{\partial \Phi(x)}{\partial x_j})$
As pointed out in comments, I need to distinguish the case $i=j$ and $i \ne j$. [Recall that in Evans $v'(|x|)=v'(r)=C r^{1-N}$ ]

So, for $i \ne j$:
$ \frac{\partial}{\partial x_i} [\frac{\partial \Phi(x)}{\partial x_j}]=\frac{\partial}{\partial x_i}[v'(|x|) \frac{x_j}{|x|}]=v''(|x|) \frac{x_i x_j}{|x|^2} - v'(|x|) \frac{x_i x_j}{|x|^3} $
Pluggin in the definition of $v(|x|)$:
\begin{align}
a_{ij}= -N \frac{x_i x_j}{|x|^{2+N}}, \quad i \ne j
\end{align}

In the case $i=j$ I have:
$\frac{\partial}{\partial x_i}[v'(|x|) \frac{x_i}{|x|}]=v''(|x|) \frac{x_i^2}{|x|^2} + v'(|x|) [ \frac{1}{|x|^2} - \frac{x_i^2}{|x|^3}]=C(1-N) \frac{x_i^2}{|x|^{N+2}}+\frac{C}{|x|^N}-C \frac{x_i^2}{|x|^{N+2}}$
So,
\begin{align}
a_{ij}=
\frac{C}{|x|^N}-NC \frac{x_i^2}{|x|^{N+2}}, \quad i =j
\end{align}

In order to compute $\star$, I use that $x_i^2 \leq |x|^2$ and so I can bound $a_{ii}$ in the following way
$a_{ii} \leq \frac{C}{|x|^N}+NC \frac{|x|^2}{|x|^{N+2}}=C(1+N) \frac{1}{|x|^N}$
So,
\begin{align}
\sum_i \sum _i a_{ii}^2 \leq N^2 C^2(1+N)^2 \frac{1}{|x|^{2N}}
\end{align}
Now, bounding again $a_{ij}$ ($i \ne j$) by using the fact that $x_i x_j \leq |x|^2$, I have
\begin{align}
\sum_i \sum_j a_{ij}^2 \leq \sum_i \sum_j (\frac{N}{|x|^{N}})^2=N(N-1)N^2 \frac{1}{|x|^{2N}} 
\end{align}
Now I put together the two sums and get
\begin{align}
\frac{N^2}{|x|^{2N}} [N(N-1)+C^2(1+N)^2]
\end{align} 
hence, by taking the square root:
$|D^2 \Phi(x)| \leq \frac{N}{|x|^N} \sqrt{N(N-1)+C^2(1+N)^2}=\tilde{C} \frac{N}{|x|^N}$
where $\tilde{C}=\sqrt{N(N-1)+C^2(N+1)}$
and the bound is proved.
